1. ## Please critique possible proof of Fermat's Last Theorem

Hello,

I'm posting this on behalf of someone else. They don't want to post it themselves for the obvious reason that this subject attracts a lot of cranks, so they don't want the embarrassment of looking like one if a flaw in this proof is found.

Personally it's been so long since I did anything but the most basic maths I can't begin to critique it. After discussing this with the proof's creator I have decided to post it on the net to get feedback, and this was the first forum I found on google.

So, over to anyone!

Originally Posted by Proof of FLT

Where we're talking about X^N+Y^N=Z^N with N>2 of course... you don't need me to set out what FLT's is...

*First point: you can divide z^n by (x-y) with x^n-1 + yx^n-2 + y^2x^n-3 +... y^n-1
*Now you can divide it again giving z^n/(x-y)^2 and the other side ix x^n-2 + 2yx^n-3 + 3y^2x^n-4 + ... with a remainder ny^n-1/(x-y)
*Now you can multiply the RHS by a factor to produce an integer and the remainder by the same amount BUT the factor is less (usually) then (x-y).
*Therefore y and z must have co factors - which contradicts initial supposition (x,y,z coprime)
*However this fails if (x-y) = r^n (r = a combination of prime factors of z)
*However there is more!
*You can represent any solution to the pytahgorean equation by 3 numbers x^2 + y^2, x^2-y^2, and 2xy
*where x and y are aribtrary coprime numbers of opposite parity.
*You can then take the 2/n th route of this
*(x^2+y^2)^2/n, (x^2-y^2)^2/n and 2xy ^2/n so that these numbers when raised to the nth power satsify the pytahgorean equation but are also of the form a^n-b^n = c^n
*Then you can factorise (x^2 + y^2)^2/n - (x^2-y^2)^2/n which must be a factor of 4x^2y^2
*If you take out 1/x^2/n you get (1 + (y^2/x^2))^2/n) - (1-(y^2/x^2))^2/n)
*You can now do a binomial expansion and extract y^2 as a common factor and also a factor of x^(2/n) which means that the exception to the rule mentioned above cannot happen.
*hence : no solutions to x^n+y^n = x^n for n = prime>2
MD5 to prove identity of proof originator on off chance this works:
e505835126f9f863215c06c9917815b6

2. Originally Posted by nondescriptname
Hello,

I'm posting this on behalf of someone else. They don't want to post it themselves for the obvious reason that this subject attracts a lot of cranks, so they don't want the embarrassment of looking like one if a flaw in this proof is found.

Personally it's been so long since I did anything but the most basic maths I can't begin to critique it. After discussing this with the proof's creator I have decided to post it on the net to get feedback, and this was the first forum I found on google.

So, over to anyone!

MD5 to prove identity of proof originator on off chance this works:
e505835126f9f863215c06c9917815b6

"First point: you can divide z^n by (x-y) with x^n-1 + yx^n-2 + y^2x^n-3 +... y^n-1" ...Uuh? Of course, we're already ruling out $x=y$ , but then

$x^n+y^n=z^n\Longrightarrow \frac{x^n+y^n}{x-y}=\frac{z^n}{x-y}$ , so what is the term above?

Certainly $x^n+y^n=(x+y)(x^{n-1}-yx^{n-2}+y^2x^{n-3}-...-y^{n-2}x+y^{n-1})$ , but this is true only if $n$ is odd, and

anyway we don't have $x-y$ , so what was intended here?

Tonio

3. Originally Posted by nondescriptname
Hello,

I'm posting this on behalf of someone else. They don't want to post it themselves for the obvious reason that this subject attracts a lot of cranks, so they don't want the embarrassment of looking like one if a flaw in this proof is found.

Personally it's been so long since I did anything but the most basic maths I can't begin to critique it. After discussing this with the proof's creator I have decided to post it on the net to get feedback, and this was the first forum I found on google.

So, over to anyone!

MD5 to prove identity of proof originator on off chance this works:
e505835126f9f863215c06c9917815b6